Proof that $\lim_{n\to\infty}{\sin{100n}}$ does not exist
How to prove that $$\lim_{n\to\infty}{\sin{100n}}$$ doesn't exist?
Some possible approaches:
It would be enough to find two subsequences $n_{k}$ that converge to two different numbers. But it's not clear how to find $n_k$ so that $\sin 100n_k$ converge.
Show that $\sin (100(n+1))-\sin 100n$ does not approach $0$. This is not obvious, either.
Solution 1:
Hint: You can use the fact that $$\sin m-\sin n=2\sin\frac{m-n}{2}\cos\frac{m+n}{2}.$$
Assume for instance that $m=n+2$ and let $n\to \infty$.
Added in Edit: If the limit existed, then of course the LHS of the above equation would be zero in limit, and so would be the RHS. This would imply that $\cos n\to 0$ which is a contradiction.